I have been reading the book Functional Equations and How to Solve Them by Christopher G. Small. In chapter 2 the first section about Cauchy's equation $f(x+y) = f(x)+f(y)$ has a theorem that states the following
Let $f : \Bbb R → \Bbb R$ satisfy Cauchy’s equation. Suppose in addition that there exists some interval $[c,d]$ of real numbers, > where $c<d$, such that $f$ is bounded from below on $[c,d]$. In other words, there exists a real number $A$ such that $f(x) ≥ A$ for all $c ≤ x ≤ d$. Then there exists a real number $a$ such that $f(x) = ax$ for all real numbers $x$.
Here continuity is not assumed. In the exercises, problem 2 is supposed to lead you through the proof of this theorem. However the problem states :
This problem asks the reader to fill in the details of the proof of Theorem 2.4, above. Let $f :\Bbb R → \Bbb R$ be a continuous function satisfying Cauchy’s equation. Suppose in addition that there exists some interval $[c, d]$ of real numbers, where $c<d$, such that $f$ is bounded from below on $[c, d]$. a) Show that $f(nx) = nf(x)$ for all real $x$. b) Define $p = d − c$. Show that $f$ is bounded from below on the interval $[0, p]$. (However, it need not be bounded below by the same constant as on the interval $[c, d]$.) c).......
My confusion here is the fact that the problem says $f$ is supposed to be continuous unlike the actual theorem also going by that logic part b of this problem is almost trivial since I just have to say that a continuous function on a closed and bounded interval must be bounded below. But then it feels like the other assumption given in this problem that the function is bounded on $[c,d]$ is completely redundant so there must be something I'm missing but I can't seem to find a way to prove part b if continuity was not given.
Sso to boil it down to what's important, my question is how to prove b if $f$ was not stated to be continuous but the assumption about boundedness was given.
Thanks in advance
Assume that $ f : \mathbb R \to \mathbb R $ is additive, and bounded below by the constant $ A $ on the non-degenerate interval $ [ c , d ] $. Then, for $ p = d - c $ and any $ x \in \mathbb R $ we have: $$ \begin {align*} & & & 0 \le x \le p \\ \implies & & & c \le x + c \le d \\ \implies & & & f ( x + c ) \ge A \\ \implies & & & f ( x ) + f ( c ) \ge A \\ \implies & & & f ( x ) \ge A - f ( c ) \end {align*} $$ Therefore $ f $ is bounded below by $ A - f ( c ) $ on $ [ 0 , p ] $.
You can see that no property of $ f $ other than additivity and boundedness on $ [ c , d ] $ is used above.